Question

If a + b + C is equal to 9 and a b + BC + CD is equal to 26 find the a square b square c square

Answer 1

$a + b + c = 9 \\ ab + bc + ca = 26$

Now,

$= > {(a + b + c)}^{2} = {(9)}^{2} \\ = > {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca) = 81 \\ = > {a}^{2} + {b}^{2} + {c}^{2} + 2(26) = 81 \\ = > {a}^{2} + {b}^{2} + {c}^{2} + 52= 81 \\ = > {a}^{2} + {b}^{2} + {c}^{2} = 81 - 52 \\ = > {a}^{2} + {b}^{2} + {c}^{2} = \pink{29}$

Answer 2

Answer:

a² + b² + c² = 29

Step-by-step explanation:

I believe your Question was,

"If a + b + c = 9 and ab + bc + ac = 26, find the a² + b² + c²."

Now we know that, using identities,

(a + b + c)² = a² + b² + c² + 2(ab + bc + ac)

Here, we know,

a + b + c = 9

ab + bc + ac = 26

Putting in the values we get,

(9)² = a² + b² + c² + 2(26)

81 = a² + b² + c² + 52

a² + b² + c² = 81 - 52

a² + b² + c² = 29

Thus,

a² + b² + c² = 29

Hope it helped and you understood it........All the best